Question

In millikan oil drop experiment a charged drop falls with a terminal velocity V. If an electric field E is applied vertically upwards it moves with terminal velocity 2V in upward direction. If electric field reduces to E/2 then its terminal velocity will be - 

(1) V/2

(2) V

(3) 3V/2 

(4) 2V 

Answer 1

Correct Answer is: (1) V/2

V = Q×E× t/m

V ∝ E

So Ans.V/2.